Question

Integrate square root of (x^2 -4)

Answer 1

you will get direct formula for sqrt(x^2-a^2) here
http://openstudy.com/users/hartnn#/updates/50960518e4b0d0275a3ccfba
or do u need to derive it ?

Answer 2

to derive, u need to use product(uv) rule of ntegration

Answer 3

see formula 23 in that link

Answer 4

\(\\ \large 23. \int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\ln| x+\sqrt{x^2-a^2}|+c
\\\)

Answer 5

So it means that what i hv 2 do is just subtitude a=2 in that rule?

Answer 6

if you can use that standard formula, then YES.
if u need to derive it, you use product rule.

Answer 7

Ok,i ll try..thx ^_^

Answer 8

if u are trying with uv rule and u don't get it, then ask.
welcome :)

Answer 9

Let U=√(x^2-4) dv= 1
du/dx= x/√(x^2-4) v=x dx
So it become:
x√(x^2-4) - ∫x^2/√(x^2-4)dx ?
then do integration again for ∫x^2/√(x^2-4) dx ...stuck T.T

Answer 10

take dv=√(x^2-4)

Answer 11

Hmm...hw should i determine which 2 put u n dv?

Answer 12

wait, which formula of uv are you using ?

Answer 13

try this
\(\\ \huge \int uv\:dx=u\int v\:dx-\int(\frac{du}{dx}\int v.dx)dx
\\\)
v=1
u=sqrt(....)
then you will also need
\(\\ \huge 19. \int \sqrt{\frac{1}{ {x^2-a^2}}}dx=\ln|x+\sqrt{x^2-a^2}|+c
\\ \)

Answer 14

\[x \sqrt{x ^{2}-4} - \int\limits \frac{ x ^{2} }{\sqrt{x ^{2}-4} }dx\]
Then for \[\int\limits \frac{ x ^{2} }{ \sqrt{x ^{2}-4} }dx\], let u=x^2 , dv= 1/ sqrt (x^2-4)
get du/dx=2x v= \[\int\limits \frac{ 1 }{ \sqrt{x ^{2}-4} }\]= \[\ln \left| x+\sqrt{x ^{2}-4} \right| +c\]
then combine them by using the rule?